Series $\frac{x^{3n}}{(3n)!}$ find sum using differentiation

Find sum of the series $$\sum_{n=1}^{\infty}\frac{x^{3n}}{\left(3n\right)!}$$ using differentiation. So far I found that $$S(x)+1=S”'(x)$$ but it does not help. Then I tried different interesting ideas like $$S(x)+S'(x)+S”(x)=e^x-1\,.$$Maybe if I get the third equation it will allow me to construct a kind of differential equation. Then, by solving it, obtain $S(x)$.

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Define $$y(x)=\sum_{k\geq0}\frac{x^{3k}}{\left(3k\right)!}.$$ We observe that $$\sum_{k\geq0}\frac{x^{3k}}{\left(3k\right)!}+\sum_{k\geq0}\frac{x^{3k+1}}{\left(3k+1\right)!}+\sum_{k\geq0}\frac{x^{3k+2}}{\left(3k+2\right)!}=e^{x}$$ so we have the second order ODE $$y”\left(x\right)+y’\left(x\right)+y\left(x\right)=e^{x}.\tag{1}$$ Let start to solve the characteristic polynomial $$\lambda^{2}+\lambda+1=0\Leftrightarrow\lambda_{1,2}=-\frac{1}{2}\pm i\frac{\sqrt{3}}{2}$$ so a solution of the homogeneous equation is $$y_{o}\left(x\right)=c_{1}e^{-x/2}\sin\left(\frac{\sqrt{3}x}{2}\right)+c_{2}e^{-x/2}\cos\left(\frac{\sqrt{3}x}{2}\right),\, c_{1},c_{2}\in\mathbb{R}.$$ Now we have to found a particolar solution. Since the known term is $e^{x}$ we know that a particular solution will be of the form $$y_{p}\left(x\right)=Ae^{x}\tag{2}$$ where $A\in\mathbb{R}$. So substituting $(2)$ in $(1)$ and equaling the coefficients, we get $$A=\frac{1}{3}$$ so, since $y\left(0\right)=1,\,y’\left(0\right)=0$ we have $$y\left(x\right)=y_{o}\left(x\right)+y_{p}\left(x\right)=\frac{2}{3}e^{-x/2}\cos\left(\frac{\sqrt{3}x}{2}\right)+\frac{e^{x}}{3}$$ and finally $$\sum_{k\geq1}\frac{x^{3k}}{\left(3k\right)!}=\frac{2}{3}e^{-x/2}\cos\left(\frac{\sqrt{3}x}{2}\right)+\frac{e^{x}}{3}-1.$$

Let $\omega=\exp\left(\frac{2\pi i}{3}\right)$. Then $\frac{\omega^n+\omega^{2n}+1}{3}$ equals $1$ iff $3\mid n$, hence:

$$\sum_{n\geq 1}\frac{x^{3n}}{(3n)!} = \color{red}{-1+\frac{e^{x}+e^{\omega x}+e^{\omega^2 x}}{3}}. \tag{1}$$

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Leftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\, #2 \,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$

\begin{align}
\color{#f00}{\sum_{n = 1}^{\infty}{x^{3n} \over \pars{3n}!}} & =
-1 + \sum_{n = 0}^{\infty}{x^{n} \over n!}\sum_{k = 0}^{\infty}\delta_{3k,n} =
-1 + \sum_{n = 0}^{\infty}{x^{n} \over n!}\sum_{k = 0}^{\infty}
\oint_{\verts{z} = a}{1 \over z^{n – 3k + 1}}\,{\dd z \over 2\pi\ic}
\end{align}

Then,
\begin{align}
\color{#f00}{\sum_{n = 1}^{\infty}{x^{3n} \over \pars{3n}!}} & =
-1 + \oint_{\verts{z} = a}\sum_{k = 0}^{\infty}z^{3k – 1}
\sum_{n = 0}^{\infty}{\pars{x/z}^{n} \over n!}\,{\dd z \over 2\pi\ic} =
-1 + \oint_{\verts{z} = a}
{\expo{x/z} \over z\pars{1 – z^{3}}}\,{\dd z \over 2\pi\ic}
\\[3mm] & =
-1 + \oint_{\verts{z} = 1/a}{z^{2}\expo{xz} \over z^{3} – 1}
\,{\dd z \over 2\pi\ic} =
$r$’s are roots of $z^{3} – 1 = 0$ whith $\verts{r} = 1 < 1/a > 1$: